New Cold Water Temperature = Wet Bulb Temperature + New Approach

Fifth Step: Compute the cold water temperature with the result of iteration as follows;

New CWT = WBT + New Approach = 80 + 10.32 = 90.32^oF
New HWT = CWT + Range = 90.32 + 12 = 102.32^oF

Through this example, it was proven that the cold water temperature at the slope of - 0.7 is slightly lower than - 0.13.

Example 4-3. The example number 18 was based on the assumption that the heat load is constant for the increase of water flow rate. Estimate the cold & hot water temperature under the assumption that the cooling range is constant for the increase of water flow rate to 20,000 from 16,000 GPM.

(Solution)
Range, R₁ = R₂ = HWT - CWT = tw₂ - tw₁ = 104 - 89 = 15^oF
Water Flow Rate in Pound, L₁ = Water Flow Rate x (500 / 60) = 16,000 x (500 / 60) = 133,333.3 lb/min
Heat Load, D₁ = L₁ x R₁ = 133,333.3 x 15 = 2,000,000 BTU/min
Air Mass Flow Rate, G₁ = G₂ = 80,848 lb/min
Liquid to Gas Ratio, L/G₁ = L₁ / G₁ = 133,333.3 / 80,848 = 1.6492
Water Flow Rate in Pound, L₂ = Water Flow Rate x (500 / 60) = 20,000 x (500 / 60) = 166,666.7 lb/min
Heat Load, D₂ = L₂ x R₂ = 166,666.7 x 15 = 2,500,000 BTU/min
Liquid to Gas Ratio, L/G₂ = L₂ / G₂ = 166,666.7 / 80,848 = 2.0615

The value of NTU at the design conditions is same as a value calculated in the example 4-1. The value of "C" of tower characteristic for the design conditions same as the example 4-1. The new tower characteristic for the increased water flow is also same as the example 4-1. Iterate until the value of new characteristic is equal to the new NTU varying the value of approach.

New Cold Water Temperature = Wet Bulb Temperature + New Approach

Fifth Step: Compute the cold water temperature with the result of iteration as follows;

New CWT = WBT + New Approach = 80 + 12.01 = 92.01^oF
New HWT = CWT + Range = 92.01 + 15 = 107.01^oF

Example 4-4. Assume again the conditions of example 4-1 and determine the cold and hot water temperature when the heat load is added to increase the cooling range from 15 to 20^oF, assuming no change in the water circulation rate or in entering air mass flow rate or wet bulb temperature.

(Solution)
Range, R₁ = HWT - CWT = tw₂ - tw₁ = 104 - 89 = 15^oF
Water Flow Rate in Pound, L₁ = L₂ = Water Flow Rate x (500 / 60) = 16,000 x (500 / 60) = 133,333.3 lb/min
Air Mass Flow Rate, G₁ = G₂ = 80,848 lb/min
Liquid to Gas Ratio, L/G₁ = L₁ / G₁ = L/G₂ = 133,333.3 / 80,848 = 1.6492
Range, R₂ = 20^oF
The value of NTU, and "C" at the design conditions is same as a value calculated in the example 4-1. Also, the new tower characteristic for even a increased cooling range is same as the example 4-1. Iterate until the value of new characteristic is equal to the new NTU varying the value of approach. (New Cold Water Temperature = Wet Bulb Temperature + New Approach)

Fifth Step: Compute the cold water temperature with the result of iteration as follows;

New CWT = WBT + New Approach = 80 + 10.65 = 90.65^oF
New HWT = CWT + Range = 90.65 + 20 = 110.65^oF

Example 4-5. Assume the existing mechanical-draft cooling tower is operating at the initial conditions of example 4-1. Determine the cold & hot water temperature if the air mass flow rate is reduced to 53,900 lb/min by the adjustment of the fan pitch angle and/or fan speed.

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